In acute $\triangle ABC$, show $DE+DF \leq BC$, where $D$, $E$, $F$ are the feet of the altitudes from $A$, $B$, $C$, respectively.

Question

Let $\triangle ABC$ be an acute angled triangle. The feet of the altitudes from $A$, $B$, and $C$ are $D$, $E$, and $F$, respectively. Prove that $$DE+DF \leq BC$$ and determine the triangles for which equality holds.

(Note: The altitude from A is the line through A which is perpendicular to BC. The foot of this altitude is the point D where it meets BC. The other altitudes are similarly defined. )

Answer 1

Due to the existence of the 9-point circle, the circumradius of the orthic triangle $DEF$ equals $\frac{R}{2}$.
Since the angles of the orthic triangle equal $\pi-2A,\pi-2B,\pi-2C$, your inequality is equivalent to $$ \sin(2B)+\sin(2C)\leq 2\sin(A) $$ or to $$ 2\sin(B)\cos(B)+2\sin(C)\cos(C)\leq 2\sin(B)\cos(C)+2\sin(C)\cos(B) $$ or to $$ (\cos B-\cos C)(\sin B-\sin C) \leq 0 $$ which is trivial since over the interval $\left(0,\frac{\pi}{2}\right)$ the sine function is increasing and the cosine function is decreasing. Equality holds only for isosceles triangles.

Answer 2

Let $F'$ be a reflection of $F$ across $BC$, so $FD = F'D$. Since $BC$ is diameter of circle around cyclic quadrilateral $BCEF$ and $\angle FDA = \angle EDA$ we see that $F'$ is on this circle and that $F',D,E$ are colinear. Since $BC$ is diameter we have $$BC\geq F'E = F'D+DE = FD+DE$$